Question 1 (a)

  • HOI is a weaker acid than HOCl

    (iii) Hypoiodous acid has the formula HOI. Predict whether HOI is a
stronger acid or a weaker acid than the acid that you identified in
part (a)(i). Justify your prediction in terms of chemical bonding. HOI
is a weaker acid than HOCI because the O—H bond in HOI is stronger
than the O—H One point is earned for predicting that HOI is a bond in
HOCI. The lower electronegativitv weaker acid than HOCI and stating
that iodine has a (electron-drawing ability) of I compared with lower
electronegativity than chlorine and EITHER that of Cl results in an
electron density that is higher (hence a bond that is stronger)
between the H and O atoms in HOI compared with the electron density
between the H and O atoms in HOCI. OR The conjugate base OCI- is more
stable than OE because Cl, being more electronegative, is better able
to accommodate the negative charge. • stating that this results in a
stronger O—H bond in HOI OR • stating that this decreases the
stability of the OE ion in solution.

  • Inductive Effects

    Atoms or groups of atoms in a molecule other than those to which H
is bonded can induce a change in the distribution of electrons within
the molecule. This is called an inductive effect, and, much like the
coordination of water to a metal ion, it can have a major effect on
the acidity or basicity of the molecule. For example, the hypohalous
acids (general formula HOX, with X representing a halogen) al have a
hydrogen atom bonded to an oxygen atom. In aqueous solution, they all
produce the following equil'brium: + OV(aq) IIOX@) The addities of
these acids vary by about three orders of magnitude, electronegativity
of the halogen atoms: HOX HOC HOBr HOI Electronegativity of X 3.0 2.8
2.5 (16.8.3) however, due to the difference in pKa 7.40 8.55 10.5 As
the electronegativity of X increases, the distribution of electron
density within the molecule changes: the electrons are drawn more
strongly toward the halogen atom and, in turn, away from the H in the
O—H bond, thus weaken•ng the O—H bond and allowing d"ssociation of
hydrogen as H +. The acidity of oxoac•ds, with the general formula
HOXOn (with n = 0—3), depends strongly on the number of terminal
oxygen atoms attached to the central atom X. As shown in Figure
16.8.1, the Ka values of the oxoac•ds of chlorine increase by a factor
of about 104 to 106 with each oxygen as successive oxygen atoms are
added. The •ncrease in acid strength with increasing number of
terminal oxygen atoms is due to both an inductive effect and increased
stabilization of the conjugate base.

    Because oxygen is the second most electronegative element, adding
terminal oxygen atoms causes electrons to be drawn away from the O—H
bond, making it weaker and thereby increasing the strength of the
acid. The colors in Figure 16.8.1 show how the electrostatic
potential, a measure of the strength of the interact•on of a point
charge at any place on the surface of the molecule, changes as the
number of terminal oxygen atoms increases. In Figure 16.8.1 and Figure
16.8.2, blue corresponds to low electron densities, while red
corresponds to high electron densities. The oxygen atom in the O—H
unit becomes steadily less red from HCIO to HC104 (aso written as
HOC103, while the H atom becomes steadily bluer, •ndicating that the
electron density' on the O—H unit decreases as the number of terminal
oxygen atoms increases. The decrease in electron density in the O—H
bond weakens it, making it easier to lose hydrogen as 11+ ions,
thereby increasing the strength of the acid.

    HOC102 (pKa = -2.0) electron rich HOC103 (pKa = -7.3 to -8) electron

Question 2 (e)

  • Percent error is always positive

    Error = Experimental Value- Accepted Value — Can be positive or
negative Percent Error = errorl/ accepted value \* 100% — or
(experimental value) - (known value) % error = known value — Always

Question 3 (e)

Step 1 Step 2 Step 3 Step 4 Step 5 C12 CH3 + HCI CH3 + C12 CH3C1 +
 Cl CH3C1 + CH2C12 + H HCI fast equilibrium slow fast fast fast

1 The order of the reaction with respect to C12 is \[Cl\]2 -z.
 \[Cl\] 1<1/2 \[Cl \]1/2 For step 1, K - \[C121 Substituting into the
 rate law for step 2 (the slowest step in the mechanism): rate-
 k\[CH4\] \[Cl\] \[C12\]l/2) \[CH4\] \[C12\]l/2 Because the exponent of
 C12 in the rate law is 1/2, the order of the reaction with respect to
 C12 is 1/2. One point is earned for the correct answer with
 appropriate justification.

  • Mechanisms with a Fast Initial Step

    Let's consider an alternative mechanism that does not involve a
termolecular step: (fast) NO(g) + Bh(g) Step 1: step 2: NOBh(g) +
NO(g) 2NOBr(g) (slow) \[14.27) In this mechanism, step 1 involves two
processes: a forward reaction and its reverse. Because step 2 is the
rate-determining step, the rate law for that step governs the rate of
the overall reaction: Rate = \[14.28) Note that NOBr, is an
intermediate generated in the forward reaction of step 1. Inter-
mediates are usually unstable and have a low, unknown concentration.
Thus, the rate law of Equation 14.28 depends on the unknown
concentration of an intermediate, which isn't desirable. We want
instead to express the rate law for a reaction in tern-IS of the
reactants, or the products if necessary, of the reaction. With the aid
of some assumptions, we can express the concentration of the inter-
mediate NOBr2 in terms of the concentrations of the starting reactants
NO and Br2. We first assume that NOBr, is unstable and does not
accumulate to any significant extent in the reaction mixture. Once
formed, NOBr2 can be consumed either by reacting with NO to form NOBr
or by falling back apart into NO and Br,. The first of these
possibili- ties is step 2 of our alternative mechanism, a slow
process. The second is the reverse of step 1, a unimolecular process:
NO(g) + NOBr2(g) Because step 2 is slow, we assume that most of the
NOBr2 falls apart according to this reaction. Thus, we have both the
forward and reverse reactions of step 1 occurring much faster than
step 2. Because they occur rapidly relative to step 2, the forward and
reverse reactions of step 1 establish an equilibrium. As in any other
dynamic equilib- rium, the rate of the forward reaction equals that of
the reverse reaction: kCNO\]CBr2\] Rate Of forward reaction CNOBr2\]
Rate Of reverse r eaction

    Solving for \[NOBr21, we have \[NOBQ = CNO\]CBr2\] Substituting this
relationship into Equation 14.28, we have Rate = = where the
experimental rate constant k equals k2k1/k\_1. This expression is
consistent with the experimental rate law (Equation 14.25). Thus, our
alternative mechanism

    (Equation 14.27), which involves two steps but only unimolecular and
bimolecu- lar processes, is far more probable than the single-step
termolecular mechanism of Equation 14.26. In general, whenever a fast
step precedes a slow one, we can solve for the concentra- tion of an
intermediate by assuming that an equilibrium is established in the
fast step.

Question 5 (d)

The total bond energy of the reactants is larger. Reaction Y is
 endothermic (AHO = +41 W mol-I > 0), so 298 there is a net input of
 energy as the reaction occurs. Thus, the total energy required to
 break the bonds in the reactants must be greater than the total energy
 released when the bonds are formed in the pro ucts. One point is
 earned for the correct answer with appropriate explanation.

Question 5 (e)

  • Thermodynamic data for an overall reaction have no bearing on how slowly or rapidly the reaction occurs

Question 6 (c)

(c) In the H2S molecule, the H—S—H bond angle is close to 900. On
 the basis of this information, which atomic orbitals of the S atom are
 involved in bonding with the H atoms? The atomic orbitals involved in
 bonding with the H atoms in H2S are p (specifically, 3p) orbitals. The
 three p orbitals are mutually perpendicular (i.e., at 900) to one
 another. One point is earned for the correct answer.

The three p orbitals are aligned along perpendicular axes

Question 6 (d)

(ii) Compare the strength of the dipole-dipole forces in liquid H2S
 to the strength of the dipole-dipole forces in liquid H20. Explain.
 The sfrength of the dipole-dipole forces in liquid H2S is weaker than
 that of the dipole-dipole forces in liquid H20. H2S molecule is less
 than that of the H20 molecule. This results from the lesser -polarity
 of the H—S bond compared with that of the H—O bond (S is less
 electronegative than O). One point is earned for the correct answer
 with a correct explanation.

  • Electronegativity and dipole-dipole force

    Permanent Dipoles • These occur when a two atoms sharing a covalent
bond have substantially different electronegativity. • HCI has a
permanent dipole as Cl has a much higher electronegativity than H. •
Molecules with a permanent dipole are described as polarmolecules

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